From an urn containing a white and b black balls, balls are successively drawn without replacement until only those of the same colour are left.

You can consider the problem as drawing balls and there is only one ball left. That's because when there are more than 1 ball left but all are the same color, it doesn't hurt to keep drawing out balls.

Also, there is the same probability of leaving any ball last.

So the probability is $\frac a{a+b}$

Its the number of ways a white ball is the last ball to be chosen divided by the total number of ways of arranging all the balls.

This is:

$$p = \frac{\frac{(a+b-1)!}{(a-1)!b!}}{\frac{(a+b)!}{a!b!}} = \frac{(a+b-1)!}{(a-1)!b!}\cdot \frac{a!b!}{(a+b)!} = \frac{a}{a+b}$$